Question

What are key differences between psychological first aid and psychotherapy? Consider both long-term and short-term

forms of psychotherapy.Using a specific example, when should crisis responders use spiritual techniques when working with those in crisis?

 

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Question

1.  Record the following for each of the three trials:
·  Trial

1

a	Mass of the empty calorimeter (g)18.6g
b	Initial temperature of the calorimeter (°C)21.5°C
c	Maximum temperature in the calorimeter from the reaction (°C)35.0°C
d	Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial35.0°C-21.5°C=13.5°C
e	Mass of the calorimeter and its contents after the reaction (g)68.74g
f	Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.74g-18.6g=50.14g
g	Calculate the moles of Mg reacted (MW = 24.305 g/mole)?50mL of 1M HCI is 50/1000×1=0.05.15/24.305=.006g/mole

·  Trial 2

a	Mass of the empty calorimeter (g)18.6g
b	Initial temperature of the calorimeter (°C)21.5°C
c	Maximum temperature in the calorimeter from the reaction (°C)44.0°C
d	Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial44.0°C-21.5°C=22.5°C
e	Mass of the calorimeter and its contents after the reaction (g)68.83g
f	Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.83-18.6=50.23
g	Calculate the moles of Mg reacted (MW = 24.305 g/mole).25/24.305=.010g/mole

·  Trial 3

a	Mass of the empty calorimeter (g)18.6g
b	Initial temperature of the calorimeter (°C)21.5°C
c	Maximum temperature in the calorimeter from the reaction (°C)52.9°C
d	Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial52.9°C-21.5°C=31.4°C
e	Mass of the calorimeter and its contents after the reaction (g)68.92g
f	Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.92g-18.6g=50.32
g	Calculate the moles of Mg reacted (MW = 24.305 g/mole).35/24.305=.014g/mole

2.  Calculate the heat released into the solution for the 3 reactions, according to the formula:

qreaction= (Ccal * ΔT) + (mcontents* Cpcontents* ΔT)

3.  
Assume Cp_contents= Cp_water= 4.18 J/g °C

a	Trial 1 (J)(4.18J/g°C)(68.74g)(35.0°C-21.5°C)=3878.9982 J/g°C
b	Trial 2 (J)(4.18J/g°C)(68.83g)(44.0°C-21.5°C)=6473.4615 J/g°C
c	Trial 3 (J)(4.18J/g°C)(68.92g)(52.9°C-21.5°C)=9045.88784 J/g°C

4.  Find the molar heat of reaction for each experiment in units of kilojoules / (mole of Mg) by dividing the heat of reaction (converted to kJ by dividing by 1000) by the moles of Mg used.

a	Trial 1 (kJ/mol)
b	Trial 2 (kJ/mol)
c	Trial 3 (kJ/mol)

5.  Calculate and record the average molar heat of reaction from the three results:

 

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